MCQ Trigonometry - Fundamentals and Basic Identities

In this problem, we can use the identities ${\displaystyle \cot <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}$ and ${\displaystyle \csc <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{1}{\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}$

${\displaystyle (1-<!--\; -\; -->{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})\cdot <!--\; \cdot \; -->\cot <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\cdot <!--\; \cdot \; -->{\csc }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=(1-<!--\; -\; -->{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})\cdot <!--\; \cdot \; -->\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}\cdot <!--\; \cdot \; -->(\frac{1}{\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{)}^2}$

${\displaystyle =(1-<!--\; -\; -->{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})\cdot <!--\; \cdot \; -->\frac{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}\cdot <!--\; \cdot \; -->\frac{1}{{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}$

This doesn't seem to help a whole lot. However, we should recognize that (1−cos²θ)=sin²θ because of the Pythagorean identity sin²θ+cos²θ = 1 .